Ram weight – Drop height
500 kN´s påle
The capacity of the pile is both a geotechnical and a
structural issue
Geotechnical CapacityThe geotechnical capacity of a driven pile
is the ability of the surrounding soil and/or the rock to withstand loads without harmful movements.
Geotechnical capacityThe rule of thumb in Sweden for driven piles
to set less than 10mm/10bl
Concrete piles: 1 kN per cm2
Steel piles: 13 kN per cm2
Geotechnical capacityOften much higher capacities is used, for
concrete piles up to 1,6kN/cm2 and steel piles 22kN/cm2. This require that the pile is driven to solid rock in order to verify capacity.
To verify capacity 5% to 25% of the piles are normally tested.
Ram weightOne condition in order to install the pile to
the desired capacity is to have sufficient ram weight. A rule of thumb in Sweden is that for micro piles it is recommended that the piston in hydraulic hammers has a weight at least 2 times the pile weight per meter.
Ram weight – Drop heightIn order to mobilize available capacity the
pile final set for the test blow has to be a few millimeters.
This require both drop height and ram weight.
Rule of thumb:
Drop height 7-8 % of the length of the pile
Ram weight 1-1,5 % of desired capacity
Ram weight- Drop heightThe relationship between impact velocity and
drop height
Kinetic Energy = Potential Energy
mv2/2 = mgH
H = v2/2g or
v = √2gH
To drop one kg on the toe from one meter is equally painful as dropping 10kg from the same height. Only the the 10kg hurts for a longer time ;-)
Drop heightExample 1: Calculation of Z (EA/c) steel pipe
pile 140mm x 8mm
A (Area): 33 cm2 (0,0033 m2)
E (Elasticitetsmodul): 210000 MPa
c (wavespeed): 5120 m/s
Z=210000*106*0,0033/5120=135000 Ns/m=
135 kNs/m
Ram weight-Drop heightIf a pile is struck by a impact velocity of 3
m/s, the force in the pile will be:
F = v*Z = 3*135 = 405 kN
To achieve this partical velocity (force) the drop height has to be:
H = v2/2g = 32/2*10 = 0,45 m
Due to loss of energy in the cap, cushioning and so forth the drop height has to be higher
Ram weight –Drop heightIf the pile is struck by an impact velocity of 6
m/s the force in the pile will be:
F = 6*135 = 810 kN
To achieve this partical velocity (force) the drop height has to be:
H = v2/2g = 62/2*10 = 1,8 m
Due to loss of energy in the cap, cushioning and so forth the drop height has to be higher
Ram weight-Drop height
Piles driven to solid bed rock the downward traveling compression wave is superimposed which results in higher stresses than that induced by the hammer (specially on short piles)
Ram weight – Drop heightThe stresses in micro piles are commonly
very close to the yeild stress for the test blow. Important that the pile is cut correct prior to test and that the hammer is lined up correctly.
Ram weight – Drop heightExample 2: Calculation of Z (EA/c) for a
concrete pile (side=235mm)
A (Area): 552 cm2 (0,0552 m2)
E (Elastic modulus): 40000 MPa
c (Wave speed): 3900 m/s
Z=40000*106*0,0552/3900=566000 Ns/m=
566 kNs/m
Ram weight – Drop heightIf the pile is struck by an impact velocity of 3
m/s the force in the pile will be:
F=3*566=1698 kN
To achieve this partical velocity (force) the drop height has to be:
H=v2/2g = 32/2*10 = 0,45 m
Due to loss of energy in the cap, cushioning and so forth the drop height has to be higher
Ram weight – Drop heightIf the pile is struck with a drop height of 1,2m
the partical velocity will be:
V= √2gH = √2*10*1,2 = 4,9 m/s
The force in the pile will than be (energy losses are neglected):
F=4,9*566=2773 kN, which is equal to a stress of 2,773/0,0552 = 50 MPa.
The pile will only withstand a few of these blows!
Ram weight – Drop heightWarning!
Modern machinery are often equiped with accelerating hammers. These hammers can quite easily over stress the pile.
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